a,\((\) 1\(-\) \(\dfrac{1}{3}\)\()\)x\((\)1\(-\)\(\dfrac{2}{5}\)\()\)x\((\)1\(-\)\(\dfrac{2}{7}\)\()\)x\((\)1\(-\)\(\dfrac{2}{9}\)\()\)
b,\(\dfrac{1}{1x3}\) + \(\dfrac{1}{3x5}\) + \(\dfrac{1}{5x7}\) + \(\dfrac{1}{7x9}\)
a) \(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times\left(1-\dfrac{2}{9}\right)\)
\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{5}{5}-\dfrac{2}{5}\right)\times\left(\dfrac{7}{7}-\dfrac{2}{7}\right)\times\left(\dfrac{9}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{2}{3}\times\dfrac{3}{5}\times\dfrac{5}{7}\times\dfrac{7}{9}\)
\(=\dfrac{2\times3\times5\times7}{3\times5\times7\times9}\)
\(=\dfrac{2}{9}\)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(=1-\dfrac{1}{9}\)
\(=\dfrac{9}{9}-\dfrac{1}{9}\)
\(=\dfrac{8}{9}\)
Tính tổng sau:
\(\dfrac{2}{3X5}+\dfrac{2}{5X7}+\dfrac{2}{7X9}+...+\dfrac{2}{19X21}\)
`2/[3.5]+2/[5.7]+2/[7.9]+....+2/[19.21]`
`=1/3-1/5+1/5-1/7+1/7-1/9+....+1/19-1/21`
`=1/3-1/21`
`=6/21`
\(1-\dfrac{2}{3x5}-\dfrac{2}{5x7}-\dfrac{2}{7x9}-...-\dfrac{2}{19x21}\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(=1-\dfrac{6}{21}=\dfrac{15}{21}=\dfrac{5}{7}\)
\(\dfrac{2}{1x3} + \dfrac{2}{3x5} + \dfrac{2}{5x7} + \dfrac{2}{7x9} + \dfrac{2}{9x11}\)
\(\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+\dfrac{2}{9x11}\)
giúp mik với ạk
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}\right)\)
\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=1-\dfrac{1}{11}\)
\(=\dfrac{11}{11}-\dfrac{1}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{2}{11}\)
\(=1-\dfrac{2}{11}\)
\(=\dfrac{9}{11}\)
\(\dfrac{2}{1x3}\)+\(\dfrac{3}{3x5}\)+\(\dfrac{2}{5x7}\)+....+\(\dfrac{2}{99x101}\)
giúp mình với ạ
`2/(1xx3)+2/(3xx5)+2/(5xx7)+...+2/(99xx101)` đề phải ntn chứ mà nhỉ
`=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101`
`=1/1-1/101`
`=101/101-1/101`
`=100/101`
(Sửa phần 3 / 3 x 5 = 2 / 3 x 5)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\)
Ta có: \(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\times\left(1-\dfrac{1}{101}\right)\)
\(=2\times\dfrac{100}{101}\)
\(=\dfrac{200}{101}\)
Sửa bài ( dòng 3 đến hết bài )
... = \(2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
so sánh A và B biết
A=\(\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{99x100}\)
B=\(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{97x99}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)
I=\(\dfrac{1}{1x3}\)+\(\dfrac{1}{3x5}\)+\(\dfrac{1}{5x7}\)+....\(\dfrac{1}{197x199}\)+\(\dfrac{1}{199x201}\)
\(I=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}\)
Lời giải:
\(2\times I=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{199\times 201}\)
\(=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+....+\frac{201-199}{199\times 201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)
\(I=\frac{200}{201}:2=\frac{100}{201}\)
K=\(\dfrac{4}{1x3}\)+\(\dfrac{4}{3x5}\)+\(\dfrac{4}{5x7}\)+...+\(\dfrac{4}{297x299}\)+\(\dfrac{4}{299x301}\)
\(K=\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+...+\dfrac{4}{299\times301}\)
\(=2\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{299\times301}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)
\(=2\times\left(1-\dfrac{1}{301}\right)=2\times\dfrac{300}{301}=\dfrac{600}{301}\)
\(K=\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+...+\dfrac{4}{299\cdot301}\)
\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)
\(=2\cdot\dfrac{300}{301}=\dfrac{600}{301}\)
